Problem: Consider the polar curve $r=\theta ^{2}$. What is the slope of the tangent line to the curve $r$ when $\theta = \dfrac{\pi}{2}$ ? Give an exact expression. $\text{slope }=$
Answer: The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={\theta^{2}}\cos(\theta) \\\\ y&={\theta^{2}} \sin(\theta) \end{aligned}$ Let's find $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)} \\\\ &=\dfrac{2\theta\sin(\theta)+\theta^2\cos(\theta)}{2\theta\cos(\theta)-\theta^2\sin(\theta)} \\\\ &=\dfrac{2\sin(\theta)+\theta\cos(\theta)}{2\cos(\theta)-\theta\sin(\theta)} \end{aligned}$ ( Note : Since we are not worried about $\theta = 0$ in this particular problem, we can divide through by $\theta$ to simplify.) Finally, we evaluate $\dfrac{dy}{dx}$ at ${\theta = \dfrac{\pi}{2}}$. $\begin{aligned}\left. \dfrac{dy}{dx} \right| _{\theta =\tfrac{\pi }{2}}&=\dfrac{2\sin\left({\dfrac{\pi}{2}}\right)+\left({\dfrac{\pi}{2}}\right)\cos\left({\dfrac{\pi}{2}}\right)}{2\cos\left({\dfrac{\pi}{2}}\right)-\left({\dfrac{\pi}{2}}\right)\sin\left({\dfrac{\pi}{2}}\right)} \\\\ &=\dfrac{2(1)+\left(\dfrac{\pi}{2}\right)(0)}{2(0)-\dfrac{\pi}{2}(1)} \\\\ &=-\dfrac{4}{\pi} \end{aligned}$ The slope of the tangent line to the curve $r$ when $\theta=\dfrac{\pi}{2}$ equals $-\dfrac{4}{\pi}$. The graph of the tangent is shown.